What is the integral of e^{-4x}cos(3x) ?

The integral of e^{-4x}cos(3x) is (3e^{-4x}sin(3x))/(25)-(4e^{-4x}cos(3x))/(25)+C

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integral of e^{-4x}cos(3x)

\int e^{- 4 x} cos (3 x) d x

\frac{3 e ^{- 4 x} sin ( 3 x )}{25} - \frac{4 e ^{- 4 x} cos ( 3 x )}{25} + C

Solution steps

\int e^{- 4 x} cos (3 x) d x

Apply Integration By Parts

: \frac{1}{3} e^{- 4 x} sin (3 x) - \int - \frac{4}{3} e^{- 4 x} sin (3 x) d x

= \frac{1}{3} e^{- 4 x} sin (3 x) - \int - \frac{4}{3} e^{- 4 x} sin (3 x) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{3} e^{- 4 x} sin (3 x) - (- \frac{4}{3} \cdot \int e^{- 4 x} sin (3 x) d x)

Apply Integration By Parts

: - \frac{1}{3} e^{- 4 x} cos (3 x) - \int \frac{4}{3} e^{- 4 x} cos (3 x) d x

= \frac{1}{3} e^{- 4 x} sin (3 x) - (- \frac{4}{3} (- \frac{1}{3} e^{- 4 x} cos (3 x) - \int \frac{4}{3} e^{- 4 x} cos (3 x) d x))

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{3} e^{- 4 x} sin (3 x) - (- \frac{4}{3} (- \frac{1}{3} e^{- 4 x} cos (3 x) - \frac{4}{3} \cdot \int e^{- 4 x} cos (3 x) d x))

Therefore

\int e^{- 4 x} cos (3 x) d x = \frac{1}{3} e^{- 4 x} sin (3 x) - (- \frac{4}{3} (- \frac{1}{3} e^{- 4 x} cos (3 x) - \frac{4}{3} \cdot \int e^{- 4 x} cos (3 x) d x))

Isolate

\int e^{- 4 x} cos (3 x) d x

= \frac{3 e ^{- 4 x} sin ( 3 x )}{25} - \frac{4 e ^{- 4 x} cos ( 3 x )}{25}

Add a constant to the solution

= \frac{3 e ^{- 4 x} sin ( 3 x )}{25} - \frac{4 e ^{- 4 x} cos ( 3 x )}{25} + C

\frac{d}{d x} (- ln (x) + x^{2})

\int_{- 6}^{8} (\frac{x}{2} + 8) d x

\int \frac{3 t ^{2}}{2} d t

\int x (ln (x))^{2} d x

\frac{d}{d x} (5 ln (2 x + 6))

What is the integral of e^{-4x}cos(3x) ?
The integral of e^{-4x}cos(3x) is (3e^{-4x}sin(3x))/(25)-(4e^{-4x}cos(3x))/(25)+C